Proof.
We only need to prove (21.12.25). By proposition 21.11.45, we can find a decreasing sequence
\[ F_ j\in \mathcal E^\infty (N_{\mathbb R},P)\cap C^\infty (N_{\mathbb R}) \]
with limit $F_\varphi $. By theorem 21.12.13, each $F_ j$ corresponds to a bounded toric potential $\varphi _ j\in \mathrm{PSH}_{\mathrm{tor}}(X,\omega )$.
By theorem 13.10.4 and theorem 21.11.55, it suffices to establish (21.12.25) for the $\varphi _ j$'s. We may therefore reduce to the case where $F_{\varphi }$ is smooth. We write $F=F_{\varphi }$ to simplify the notations. The notations $a_ i=\log |z_ i|^2$ will be used, where $i=1,\ldots ,n$.
Next we fix an identification $N=\mathbb {Z}^ n$. Fix a test function $f\in C^{0}_ c(N_{\mathbb {R}})$. We need to show that
\[ \int _{\mathbb {C}^{*n}}f(a_1,\ldots ,a_ n) \left(\mathrm{dd}^{\mathrm{c}}\operatorname{Trop}^* F(z_1,\ldots ,z_ n)\right)^ n=\int _{\mathbb {R}^ n} f\, \operatorname{MA}_{\mathbb {R}}(F). \]
Using proposition 21.11.52 and (21.12.8), this reduces to
E34E6BC222DC
21.12.26
\begin{equation} \begin{split} \left( \frac{\mathrm{i}}{2\pi }\right)^ n\int _{\mathbb {C}^{*n}}f(a_1,\ldots ,a_ n) \left(\sum _{i,j=1}^ n \partial _{i,j}F(a_1,\ldots ,a_ n)z_ i^{-1}\overline{z_ j}^{-1}\mathrm{d}z_ i\wedge \mathrm{d}\overline{z_ j}\right)^ n=\\ n!\int _{\mathbb {R}^ n} f \det \nabla ^2 F\, \mathrm{d}\operatorname{vol}. \end{split} \end{equation}
Expanding the bracket, we get
\[ \begin{aligned} \left(\sum _{i,j=1}^ n \partial _{i,j}Fz_ i^{-1}\overline{z_ j}^{-1}\mathrm{d}z_ i\wedge \mathrm{d}\overline{z_ j}\right)^ n =& \sum _{i_1,\ldots ,i_ n=1}^ n \sum _{j_1,\ldots ,j_ n=1}^ n \partial _{i_1j_1}F\cdots \partial _{i_ nj_ n}F\cdot \\ & \mathrm{d}\log z_{i_1}\wedge \mathrm{d}\log \overline{z_{j_1}}\wedge \cdots \wedge \mathrm{d}\log z_{i_ n}\wedge \mathrm{d}\log \overline{z_{j_ n}}, \end{aligned} \]
where $\mathrm{d}\log z_ i=z_ i^{-1}\mathrm{d}z_ i$ and $\mathrm{d}\log \overline{z_ i}=\overline{z_ i}^{-1}\mathrm{d}\overline{z_ i}$ are understood.
Using the apparent symmetry, the expression on the right-hand side becomes
\[ \begin{aligned} & \sum _{\sigma ,\tau \in \mathfrak {S}_ n}\prod _{k=1}^ n \partial _{\sigma (k)\tau (k)}F\, \mathrm{d}\log z_{\sigma (1)}\wedge \mathrm{d}\log \overline{z_{\tau (1)}}\wedge \cdots \wedge \mathrm{d}\log z_{\sigma (n)}\wedge \mathrm{d}\log \overline{z_{\tau (n)}},\\ =& n!\sum _{\tau \in \mathfrak {S}_ n}\prod _{k=1}^ n \partial _{k\tau (k)}F\, \mathrm{d}\log z_{1}\wedge \mathrm{d}\log \overline{z_{\tau (1)}}\wedge \cdots \wedge \mathrm{d}\log z_{n}\wedge \mathrm{d}\log \overline{z_{\tau (n)}}\\ =& n!\sum _{\tau \in \mathfrak {S}_ n}\operatorname {sgn}(\tau )\prod _{k=1}^ n \partial _{k\tau (k)}F\, \mathrm{d}\log z_{1}\wedge \mathrm{d}\log \overline{z_1}\wedge \cdots \wedge \mathrm{d}\log z_{n}\wedge \mathrm{d}\log \overline{z_ n}\\ =& n! \det \nabla ^2 F\mathrm{d}\log z_{1}\wedge \mathrm{d}\log \overline{z_1}\wedge \cdots \wedge \mathrm{d}\log z_{n}\wedge \mathrm{d}\log \overline{z_ n}, \end{aligned} \]
where $\mathfrak {S}_ n$ is the permutation group on $\{ 1,\ldots ,n\} $ and $\operatorname {sgn}(\tau )$ is the sign of $\tau $.
Next, switch to polar coordinates for each $z_ i$: Let $z_ i=r_ i\exp (\mathrm{i}\theta _ i)$ and recall that $r_ i=\exp (a_ i/2)$. Then the left-hand side of (21.12.26) becomes
\[ \begin{split} & \frac{n!}{(2\pi )^ n}\int _{\mathbb {R}^ n\times [0,2\pi )^ n} f \det \nabla ^2 F \, \mathrm{d}a_1\wedge \mathrm{d}\theta _1\wedge \cdots \wedge \mathrm{d}a_ n\wedge \mathrm{d}\theta _ n\\ =& n!\int _{\mathbb {R}^ n}f\det \nabla ^2 F \, \mathrm{d}a_1\wedge \cdots \wedge \mathrm{d}a_ n, \end{split} \]
which is exactly what we have expected.
$\square$
Pome
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